How many ways are there to seat 7 people around a round table, if we regard two arrangements that differ by a rotation as the same?

Question

I have to use Burnside's Lemma for this.

What I was thinking is that the group that acts on the set of possible seating arrangements was the rotation group $R_7$.

For the total combinations we have $7!$ since we have 7 persons.

The identity fixes every arrangement so $\mathrm{Fix}(e) = 7!$ Since every person is distinct, every other element of the action does not fix the arrangement, since you would need the heptagon of people to be composed of the same person (like the problems about colored beads on necklaces), therefore $\mathrm{Fix}(r^k) = 0$ where $r \in R_7 \setminus \{e\} $

This covers all the elements of $R_7$, so we end up with

$$\frac{7!}{7} = 6!$$ possible seating arrangements that don't differ by some rotation.

Answer 1

For cicular permutation suppose to fix an element in a circle, then you can permute the others and thus we obtain $(n-1)!$ arrangements.

Answer 2

I know nothing about the theorem you mention, but just as if you cut a necklace at a point and count the permutations and then consider the repetitions. Since for "any" given four people in a row, say

$$\color{red}{A}BCD,$$

you can also find $\color{red}BCD\color{blue}{A}$, $\color{red}{C}D\color{blue}{AB},$ $\color{red}{D}\color{blue}{ABC}$, so you have to divide it by

$$(\textrm{number of people}),$$

in short

$$\frac{(\textrm{number of people})!}{\textrm{number of people}}$$